Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = (int)110 + 3;
void tryRow(vector<vector<int> > ar, int& ans, vector<int>& tp,
            vector<int>& val) {
  int n = (int((ar).size()));
  int m = (int((ar[0]).size()));
  for (int(r) = (0); (r) < (n); (r)++) {
    int minv = ar[r][0];
    for (int(c) = (0); (c) < (m); (c)++) {
      minv = min(minv, ar[r][c]);
    }
    for (int(c) = (0); (c) < (m); (c)++) {
      ar[r][c] -= minv;
    }
    if (minv > 0) {
      while (minv--) {
        tp.push_back(0);
        val.push_back(r);
      }
    }
  }
  for (int(c) = (0); (c) < (m); (c)++) {
    int minv = ar[0][c];
    for (int(r) = (0); (r) < (n); (r)++) {
      minv = min(minv, ar[r][c]);
    }
    for (int(r) = (0); (r) < (n); (r)++) {
      ar[r][c] -= minv;
    }
    if (minv > 0) {
      while (minv--) {
        tp.push_back(1);
        val.push_back(c);
      }
    }
  }
  bool was = 0;
  for (int(r) = (0); (r) < (n); (r)++) {
    for (int(c) = (0); (c) < (m); (c)++) {
      if (ar[r][c] != 0) {
        was = true;
      }
    }
  }
  if (was) {
    ans = -1;
    return;
  }
  ans = (int((tp).size()));
}
void tryCol(vector<vector<int> > ar, int& ans, vector<int>& tp,
            vector<int>& val) {
  int n = (int((ar).size()));
  int m = (int((ar[0]).size()));
  for (int(c) = (0); (c) < (m); (c)++) {
    int minv = ar[0][c];
    for (int(r) = (0); (r) < (n); (r)++) {
      minv = min(minv, ar[r][c]);
    }
    for (int(r) = (0); (r) < (n); (r)++) {
      ar[r][c] -= minv;
    }
    if (minv > 0) {
      while (minv--) {
        tp.push_back(1);
        val.push_back(c);
      }
    }
  }
  for (int(r) = (0); (r) < (n); (r)++) {
    int minv = ar[r][0];
    for (int(c) = (0); (c) < (m); (c)++) {
      minv = min(minv, ar[r][c]);
    }
    for (int(c) = (0); (c) < (m); (c)++) {
      ar[r][c] -= minv;
    }
    if (minv > 0) {
      while (minv--) {
        tp.push_back(0);
        val.push_back(r);
      }
    }
  }
  bool was = 0;
  for (int(r) = (0); (r) < (n); (r)++) {
    for (int(c) = (0); (c) < (m); (c)++) {
      if (ar[r][c] != 0) {
        was = true;
      }
    }
  }
  if (was) {
    ans = -1;
    return;
  }
  ans = (int((tp).size()));
}
int main(int argc, char* argv[]) {
  ios::sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  vector<vector<int> > ar(n, vector<int>(m));
  for (int(i) = (0); (i) < (n); (i)++) {
    for (int(j) = (0); (j) < (m); (j)++) {
      cin >> ar[i][j];
    }
  }
  int ans1, ans2;
  vector<int> tp1, tp2, val1, val2;
  tryRow(ar, ans1, tp1, val1);
  tryCol(ar, ans2, tp2, val2);
  if (ans1 == -1) {
    cout << -1 << endl;
    return 0;
  }
  int ans;
  vector<int> tp, val;
  if (ans1 < ans2) {
    ans = ans1;
    tp = tp1;
    val = val1;
  } else {
    ans = ans2;
    tp = tp2;
    val = val2;
  }
  cout << (int((tp).size())) << endl;
  for (int(i) = (0); (i) < ((int((tp).size()))); (i)++) {
    if (tp[i] == 0) {
      cout << "row " << val[i] + 1 << endl;
    } else {
      cout << "col " << val[i] + 1 << endl;
    }
  }
  return 0;
}